Giải pt sinx.sin2x.sin3x = 1/4sin4x

Đáp án:

$\begin{array}{l}
\sin x.\sin 2x.\sin 3x = \dfrac{1}{4}\sin 4x\\
 \Rightarrow \sin x.\sin 2x.\sin 3x = \dfrac{1}{4}.2\sin 2x.cos2x\\
 \Rightarrow \sin 2x\left( {\sin x.sin3x - \dfrac{1}{2}.\cos 2x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\dfrac{{ - 1}}{2}.\left( {\cos 4x - cos2x} \right) - \dfrac{1}{2}\cos 2x = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
 - \cos 4x + \cos 2x - \cos 2x = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x = k\pi \\
\cos 4x = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
4x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}
\end{array} \right.
\end{array}$